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  • Machine Learning CodeForces

    You come home and fell some unpleasant smell. Where is it coming from?

    You are given an array a. You have to answer the following queries:

    1. You are given two integers l and r. Let ci be the number of occurrences of i inal: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, ..., c109}
    2. You are given two integers p to x. Change ap to x.

    The Mex of a multiset of numbers is the smallest non-negative integer not in the set.

    Note that in this problem all elements of a are positive, which means that c0 = 0 and 0 is never the answer for the query of the second type.

    Input

    The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.

    The second line of input contains n integers — a1a2..., an (1 ≤ ai ≤ 109).

    Each of the next q lines describes a single query.

    The first type of query is described by three integers ti = 1, liri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.

    The second type of query is described by three integers ti = 2, pixi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.

    Output

    For each query of the first type output a single integer  — the Mex of{c0, c1, ..., c109}.

    Example

    Input
    10 4
    1 2 3 1 1 2 2 2 9 9
    1 1 1
    1 2 8
    2 7 1
    1 2 8
    Output
    2
    3
    2

    Note

    The subarray of the first query consists of the single element — 1.

    The subarray of the second query consists of four 2s, one 3 and two 1s.

    The subarray of the fourth query consists of three 1s, three 2s and one 3.

    注意离散化 ,更换点也要加入数组里面进行离散化

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const int maxn = 2e5 + 10;
     5 int n, m, tim, L, R, tot, sz, qsz;
     6 int sum[maxn], cnt[maxn], ans[maxn], now[maxn];
     7 int a[maxn], b[maxn];
     8 struct node {
     9     int l, r, id, t;
    10     node() {}
    11     node(int l, int r, int id, int t): l(l), r(r), id(id), t(t) {}
    12 } qu[maxn];
    13 struct node1 {
    14     int pos, x, y;
    15     node1() {}
    16     node1(int pos, int x, int y): pos(pos), x(x), y(y) {}
    17 } c[maxn];
    18 int cmp(node a, node b) {
    19     if (a.l / sz == b.l / sz) {
    20         if (a.r / sz == b.r / sz) return a.t < b.t;
    21         return a.r < b.r;
    22     }
    23     return a.l < b.l;
    24 }
    25 void add(int val) {
    26     cnt[sum[val]]--;
    27     sum[val]++;
    28     cnt[sum[val]]++;
    29 }
    30 void del(int val) {
    31     cnt[sum[val]]--;
    32     sum[val]--;
    33     cnt[sum[val]]++;
    34 }
    35 void change(int pos, int x) {
    36     if (L <= pos && pos <= R) {
    37         del(now[pos]);
    38         add(x);
    39     }
    40     now[pos] = x;
    41 }
    42 int main() {
    43     scanf("%d%d", &n, &m);
    44     sz = (int)pow(n, 0.66666667);
    45     for (int i = 1 ; i <= n ; i++) {
    46         scanf("%d", &a[i]);
    47         b[++tot] = a[i];
    48         now[i] = a[i];
    49     }
    50     qsz = tim = 0;
    51     for (int i = 1 ; i <= m ; i++) {
    52         int op;
    53         scanf("%d", &op);
    54         if (op == 1) {
    55             int l, r;
    56             scanf("%d%d", &l, &r);
    57             qu[++qsz] = node(l, r, qsz, tim);
    58         } else {
    59             int pos, x;
    60             scanf("%d%d", &pos, &x);
    61             b[++tot] = x;
    62             c[++tim] = node1(pos, x, now[pos]);
    63             now[pos] = x;
    64         }
    65     }
    66     sort(qu + 1, qu + qsz + 1, cmp);
    67     sort(b + 1, b + tot + 1);
    68     tot = unique(b + 1, b + tot + 1) - b;
    69     for (int i = 1 ; i <= n ; i++)
    70         now[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
    71     for (int i = 1 ; i <= tim ; i++) {
    72         c[i].x = lower_bound(b + 1, b + tot + 1, c[i].x) - b;
    73         c[i].y = lower_bound(b + 1, b + tot + 1, c[i].y) - b;
    74     }
    75     tim = 0;
    76     for (int i = 1 ; i <= qsz ; i++) {
    77         while(L > qu[i].l) add(now[--L]);
    78         while(R < qu[i].r) add(now[++R]);
    79         while(L < qu[i].l) del(now[L++]);
    80         while(R > qu[i].r) del(now[R--]);
    81         while(tim < qu[i].t) tim++, change(c[tim].pos, c[tim].x);
    82         while(tim > qu[i].t) change(c[tim].pos, c[tim].y), tim--;
    83         int mex = 1;
    84         while(cnt[mex] > 0) mex++;
    85         ans[qu[i].id] = mex;
    86     }
    87     for (int i = 1 ; i <= qsz ; i++ )
    88         printf("%d
    ", ans[i]);
    89     return 0;
    90 }
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  • 原文地址:https://www.cnblogs.com/qldabiaoge/p/9359174.html
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